Conditional expectations, continuous random variables

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Комментарии
Автор

Thank you so much for your lecture.. I was almost crazy since I kept failed to find right CEF. Now
I can finally finish my assignment after watching your video. Thank you so much😭😭😭😭😭😭😭😭😭love from
Korea💙💙

qbrpupo
Автор

You are a lifesaver, Especially the part about conditional expectations and their dependence.

Question: At 6:20, why is the integral over dy and not dx?

advaitathreya
Автор

Last part about f_z(z)...I'm not sure about my step 0
2 step procedure: 1)find CDF of Z. 2)differentiate to get PDF of Z

-1) find f_X. Just integrate the joint (f(x, y) = x+y) with respect to y...or, by symmetry it's similar to the f_Y in the video. f_X = x +1/2

0) finding the inverse
From (3X+2)/(6X+3) <= z, make X the subject
3X+2 <= 6zX+3z
2-3z <= (6z-3)X
Assuming 6z-3 < 0 (i.e. z < 1/2)
(2-3z) / (6z-3) >= X (flip the inequality bcos i divided by a negative quantity)


1) CDF of Z
= F_z
= P(Z<= z)
= P((3X+2)/(6X+3) <= z)
= P(X <= (2-3z) / (6z-3)) (from step 0)
= F_X ((2-3z) / (6z-3))
= CDF of X, but evaluated at (2-3z) / (6z-3)

2) PDF of z
= d/dz [ F_X ((2-3z) / (6z-3)) ]
= f_X ((2-3z) / (6z-3)) * d/dz [ (2-3z) / (6z-3) ]
= ((2-3z) / (6z-3) + 1/2 ) * d/dz [ (2-3z) / (6z-3) ]
=...
=-1/18(2z-1)^3

But..am I allowed to assume that z<1/2 in step 0 ?

khbye